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1

Quiz-6 / Re: Q6 TUT 5301

« on: November 17, 2018, 05:34:28 PM »

Why 2k? It should just be k.

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \pi \ cot\ ( \pi z)\\
\\
=\frac{\pi cos( \pi z)}{sin( \pi z)}
\end{array}$

So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.

You are right , it was a rudimentary mistake. I have corrected my answer thanks

2

Quiz-6 / Re: Q6 TUT 5301

« on: November 17, 2018, 05:19:25 PM »

To begin, $f(z) = \pi \text{cot}(\pi z) = \pi \frac{\text{cos}(\pi z)}{\text{sin}(\pi z)}$

Let the numerator be denoted by $h(z) = \pi \text{cos}(\pi z)$,

and the denominator be $g(z) = \text{sin}(\pi z)$

Then the denominator $g(z) = \text{sin}(\pi z) = 0 \implies z = k, k \in \mathbb{Z}$

Now, $h(k) = \pi \text{cos}(k\pi) = \pi \text{ or } -\pi \neq 0 \implies \text{order = 0}$

$g(k) = \text{sin}(k\pi) = 0$

$g'(z)= \pi \text{cos}(\pi z) \implies g'(k) = \pi \text{cos}(k\pi) = \pi \text{ or } -\pi  \neq 0 \implies \text{order = 1}$

$\therefore$ this function has a pole singularity with order of pole = 1 - 0 = 1